In triangle $ABC$, we have $\angle A = 90^\circ$ and $\sin B = \frac{4}{7}$.  Find $\cos C$.
[asy]

pair A,B,C;

A = (0,0);

B = (4,0);

C = (0,sqrt(33));

draw(A--B--C--A);

draw(rightanglemark(B,A,C,10));

label("$A$",A,SW);

label("$B$",B,SE);

label("$C$",C,N);

[/asy]

Since $\triangle ABC$ is a right triangle, we have $\sin B = \frac{AC}{BC}$ and $\cos C = \frac{AC}{BC}$, so $\cos C = \sin B = \boxed{\frac47}$.